PHYS 1443 – Section 001
Lecture #9
Monday, February 28, 2011
Dr. Jaehoon Yu
•
•
•
•
Uniform Circular Motion
Motion Under Resistive Forces
Newton’s Law of Universal Gravitation
Free Fall Acceleration
Today’s homework is homework #6, due 10pm, Friday, Mar. 11!!
Announcements
• Mid-term comprehensive exam
– 19% of the total
– 1 – 2:20pm, Monday, Mar. 7, SH103
– Covers: CH.1.1 through what we complete this
Wednesday, Mar. 2 (Ch6?) plus Appendices A and B
– Mixture of multiple choices and free response problems
• Please be sure to explain as much as possible
• Just the answers in free response problems are not
accepted.
Monday, Feb. 28, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
2
Special Project
• Derive the formula for the gravitational
acceleration (gin) at the radius Rin   RE  from
the center, inside of the Earth. (10 points)
• Compute the fractional magnitude of the
gravitational acceleration 1km and 500km
inside the surface of the Earth with respect to
that on the surface. (6 points, 3 points each)
• Due at the beginning of the class Wednesday,
Mar. 9
Monday, Feb. 28, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
3
Special Project
• Two protons are separated by 1m.
– Compute the gravitational force (FG) between the two
protons (3 points)
– Compute the electric force (FE) between the two
protons (3 points)
– Compute the ratio of FG/FE (3 points) and explain
what this tells you (1 point)
• Due: Beginning of the class, Wednesday, Mar. 23
Monday, Feb. 28, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
4
Definition of the Uniform Circular Motion
Uniform circular motion is the motion of an object
traveling at a constant speed on a circular path.
Monday, Feb. 28, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
5
Speed of a uniform circular motion?
Let T be the period of this motion, the time it takes for the object
to travel once around the complete circle whose radius is r is
r
Monday, Feb. 28, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
distance
v
time
2 r

T
6
Ex. : A Tire-Balancing Machine
The wheel of a car has a radius of 0.29m and is being rotated at
830 revolutions per minute on a tire-balancing machine.
Determine the speed at which the outer edge of the wheel is
moving.
1
3
1.2

10
min revolution

830revolutions min
T  1.2  103 min  0.072 s


2 r 2 0.29 m
v

 25m s
T
0.072 s
Monday, Feb. 28, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
7
Centripetal Acceleration
In uniform circular motion, the speed is constant, but the direction
of the velocity vector is not constant.
    90
   90
   0
 
Monday, Feb. 28, 2011
The change of direction of the velocity is the same
PHYSas
1443-001,
Spring 2011
8
the change
of the angle in the circular motion!
Dr. Jaehoon Yu
Centripetal Acceleration
From the
geometry
v 2 vt 2
2

v
r
sin
r
ac
r
ac
v v

t r
2
2
What is the direction of ac?
Monday, Feb. 28, 2011
Always toward the center of circle!
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
v
ac 
r
Centripetal Acceleration
9
Newton’s Second Law & Uniform Circular Motion
The centripetal * acceleration is always perpendicular to the
velocity vector, v, and points to the center of the axis (radial
direction) in a uniform circular motion.
ar 
v2
r
Are there forces in this motion? If so, what do they do?
The force that causes the centripetal acceleration
acts toward the center of the circular path and
causes the change in the direction of the velocity
vector. This force is called the centripetal force.
v2
 Fr  mar  m r
What do you think will happen to the ball if the string that holds the ball breaks?
The external force no longer exist. Therefore, based on Newton’s 1st law,
the ball will continue its motion without changing its velocity and will fly
away along the tangential direction to the circle.
*Mirriam Webster: Proceeding or acting in the direction toward the center or axis
Monday, Feb. 28, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
10
Ex. 5.11 of Uniform Circular Motion
A ball of mass 0.500kg is attached to the end of a 1.50m long cord. The ball is moving
in a horizontal circle. If the string can withstand the maximum tension of 50.0 N, what
is the maximum speed the ball can attain before the cord breaks?
Centripetal
acceleration:
When does the
string break?
v2
ar 
r
v2
 Fr  mar  m r  T
when the required centripetal force is greater than the sustainable tension.
v2
m
 T
r
v  Tr  50.0 1.5  12.2  m / s 
m
0.500
Calculate the tension of the cord
when speed of the ball is 5.00m/s.
Monday, Feb. 28, 2011
v2
 5.00   8.33 N
 0.500 
T m
 
r
1.5
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
2
11
Example 5.15: Banked Highway
(a) For a car traveling with speed v around a curve of radius r, determine the formula
for the angle at which the road should be banked so that no friction is required to
keep the car from skidding.
y
x
2
mv
x comp.  Fx  FN sin   mar 
r
mv 2
FN sin  
r
FN cos   mg
y comp.  Fy  FN cos   mg  0
mg
FN 
cos 
2
mg sin 
mv
FN sin  
 mg tan  
cos 
r
v2
tan  
gr
(b) What is this angle for an expressway off-ramp curve of radius 50m at a design
speed of 50km/h?
v  50km / hr  14m / s
Monday, Feb. 28, 2011
tan  
142
50  9.8
 0.4
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
  tan 1 0.4  22o
12
Forces in Non-uniform Circular Motion
The object has both tangential and radial accelerations.
What does this statement mean?
Fr
F
Ft
The object is moving under both tangential and
radial forces.
ur ur
ur
F  Fr  Ft
These forces cause not only the velocity but also the speed of the ball
to change. The object undergoes a curved motion in the absence of
constraints, such as a string.
What is the magnitude of the net acceleration?
Monday, Feb. 28, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
a  ar2  at2
13
Ex. 5.12 for Non-Uniform Circular Motion
A ball of mass m is attached to the end of a cord of length R. The ball is moving in a
vertical circle. Determine the tension of the cord at any instance in which the speed
of the ball is v and the cord makes an angle θ with vertical.
V

T
R
m
What are the forces involved in this motion?
•The gravitational force Fg
•The radial force, T, providing the tension.
Fg=mg
tangential
comp.
Radial
comp.
F
t
 mg sin   mat
2
v
 Fr  T  mg cos  mar  m R
at  g sin 
 v2

T  m  g cos  
R

At what angles the tension becomes the maximum
and the minimum. What are the tensions?
Monday, Feb. 28, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
14
Motion in Resistive Forces
Medium can exert resistive forces on an object moving through it due
to viscosity or other types frictional properties of the medium.
Some examples?
Air resistance, viscous force of liquid, etc
These forces are exerted on moving objects in opposite direction of
the movement.
These forces are proportional to such factors as speed. They almost
always increase with increasing speed.
Two different cases of proportionality:
1. Forces linearly proportional to speed:
Slowly moving or very small objects
2. Forces proportional to square of speed:
Large objects w/ reasonable speed
Monday, Feb. 28, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
15
Newton’s Law of Universal Gravitation
People have been very curious about the stars in the sky, making
observations for a long time. The data people collected, however, have
not been explained until Newton has discovered the law of gravitation.
Every object in the universe attracts every other object with a force that
is directly proportional to the product of their masses and inversely
proportional to the square of the distance between them.
How would you write this
law mathematically?
G is the universal gravitational
constant, and its value is
m1 m2
Fg  2
r12
With G
G  6.673 10
11
m1m2
Fg  G
r122
Unit?
N  m2 / kg 2
This constant is not given by the theory but must be measured by experiments.
This form of forces is known as the inverse-square law, because the magnitude of the
force is inversely proportional to the square of the distances between the objects.
Monday, Feb. 28, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
16
Free Fall Acceleration & Gravitational Force
The weight of an object with mass m
is mg. Using the force exerting on a
particle of mass m on the surface of
the Earth, one can obtain
What would the gravitational
acceleration be if the object is at
an altitude h above the surface of
the Earth?
mg
g
M Em
RE2
ME
G
RE2
G
M Em  G M Em
Fg  mg '  G
2
2


R

h
r
E
ME
g'  G
from the
RE  h 2 Distance
center of the Earth
What do these tell us about the gravitational acceleration?
to the object at the
altitude h.
•The gravitational acceleration is independent of the mass of the object
•The gravitational acceleration decreases as the altitude increases
•If the distance from the surface of the Earth gets infinitely large, the weight of the
object approaches 0.
Monday, Feb. 28, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
17
Ex. 6.2 for Gravitational Force
The international space station is designed to operate at an altitude of 350km. Its designed
weight (measured on the surface of the Earth) is 4.22x106N. What is its weight in its orbit?
The total weight of the station on the surface of the Earth is
FGE  mg
ME
M Em
6
G
2  4.22  10 N
RE
Since the orbit is at 350km above the surface of the Earth,
the gravitational force at that altitude is
FO  mg '  G
MEm
RE  h 
Therefore the weight in the orbit is
FO

2
E
R
RE  h 
2
FGE 
Monday, Feb. 28, 2011

6.37  10 6
6.37  10
6
2


RE2
RE  h 
2
FGE
2
 3.50  10

5 2
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
 4.22  10 6  3.80  10 6 N
18
Example for Universal Gravitation
Using the fact that g=9.80m/s2 on the Earth’s surface, find the average density of the Earth.
Since the gravitational acceleration is
Fg
G
M Em
RE2
 mg
Solving for g
Solving for ME
Therefore the
density of the
Earth is
g
ME
M
 G 2  6.67 1011 E2
RE
RE
RE 2 g
ME 
G
2

ME

VE
RE g
3g
G


4GRE
4
3
RE

3  9.80
3
3


5
.
50

10
kg
/
m
4  6.67 10 11  6.37 106
Monday, Feb. 28, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
19